Question

我收到“表创建成功”消息但没有创建表。

<html>
        <body>
        <?php

        $con=mysql_connect("localhost","root","");

        //create db
        mysql_query("CREATE DATABASE VashDedomenwn2", $con);
        echo "Db creation success <br>";


        //create table
        $sql= mysql_query("CREATE TABLE VashDedomenwn2.phone_book
        (
        personID int NOT NULL,
        PRIMARY KEY(person ID),
        LastName varchar(20) NOT NULL,
        FirstName varchar(20),
        Address varchar(30),
        Age int,
        Phone varchar(10)
        )
        ");

        mysql_query($sql,$con);
        echo "Table creation success <br>";

        //END CONNECTION
        mysql_close($con);
        ?>

        </body>
    </html>

我是php的新手！可能这是一个愚蠢的错误...感谢帮助

Answer 1

在创建表之前，必须打开您创建的数据库。

//create db
mysql_query("CREATE DATABASE VashDedomenwn2", $con);
echo "Db creation success <br>";

mysql_select_db("database name", $con);

//create table
...

并改变

$sql= mysql_query("CREATE TABLE VashDedomenwn2.phone_book
(
...
)
");

到

$sql= mysql_query("CREATE TABLE `VashDedomenwn2.phone_book`
(
...    
)
");

好的，正如@php_purest所说，使用mysqli更好。 Mysqli是支持OOPS的改进版mysql。它可以降低服务器的压力。

像这样：

$conn = new mysqli("localhost","root","");

$conn->query("CREATE DATABASE VashDedomenwn2");
echo "Db creation success <br>";
$conn->select_db("VashDedomenwn2");

//create table
$conn->query("CREATE TABLE `VashDedomenwn2.phone_book`
(
personID int NOT NULL PRIMARY KEY,
LastName varchar(20) NOT NULL,
FirstName varchar(20),
Address varchar(30),
Age int,
Phone varchar(10)
)
");

echo "Table creation success <br>";

//END CONNECTION
$conn->close();

Javascript图像滑块不起作用

1 个答案: